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Geometry Difficulty 5.4 AIME, harder Find the answer

89. For ABC\triangle A B C, the circumradius is RR, the perimeter is PP, and the area is KK. Determine the maximum value of KPR3\frac{K P}{R^{3}}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

89. By the Law of Sines, P=a+b+c=2R(sinA+sinB+sinC),K=absinC2=2R2sinAsinBsinCP=a+b+c=2 R(\sin A+\sin B+\sin C), K=\frac{a b \sin C}{2}=2 R^{2} \sin A \sin B \sin C.

Therefore, by the AM-GM inequality,
KPR3=4sinAsinBsinC(sinA+sinB+sinC)427(sinA+sinB+sinC)4\begin{aligned} \frac{K P}{R^{3}}= & 4 \sin A \sin B \sin C(\sin A+\sin B+\sin C) \leqslant \\ & \frac{4}{27}(\sin A+\sin B+\sin C)^{4} \end{aligned}

By Jensen's inequality, sinA+sinB+sinC3sinA+B+C3=332\sin A+\sin B+\sin C \leqslant 3 \sin \frac{A+B+C}{3}=\frac{3 \sqrt{3}}{2}, so KPR3274\frac{K P}{R^{3}} \leqslant \frac{27}{4}. Equality holds if and only if ABC\triangle A B C is an equilateral triangle.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.